
Find the exact value of tan(ab) sin a = 4/5, 3pi/2<a<pi; tan b = sqrt2, pi/2<b<pi identity used is: tan(ab)=(tan atan b)/1+tan a tan b simplify answer using radicals. (a is alpha, b is beta)

I'm having a hard time understanding how to do Integrals involving tan^2. I have two problems: 1. Find the integral of (tan^2 y +1)dy 2. Find the integral of (7tan^2 u +15)du 1. My approach to it is to replace the tan^2 y portion of the problem with sec^2

Intergrate ¡ì sec^3(x) dx could anybody please check this answer. are the steps correct? thanks. = ¡ì sec x d tan x = sec x tan x  ¡ì tan x d sec x = sec x tan x  ¡ì sec x tan^2(x) dx = sec x tan x + ¡ì sec x dx  ¡ì sec^3(x) dx = sec x tan x

intergral of (x^3 4x + 3)/(2x) dx would this be ln abs(x^34x+3) + C? I don't really understand how to solve this problem. d/dx (tan (x^2)) sec^2(x^2)(2x) would this be the correct answer to find the derivative of tan (x^2)? no, because if you

Find tan(3 theta) in terms of tan theta Use the formula tan (a + b) = (tan a + tan b)/[1  tan a tan b) in two steps. First, let a = b = theta and get a formula for tan (2 theta). tan (2 theta) = 2 tan theta/[(1  tan theta)^2] Then write down the equation


Use integration by parts to evaluate the integral of x*sec^2(3x). My answer is ([x*tan(3x)]/3)[ln(sec(3x))/9] but it's incorrect. u=x dv=sec^2(3x)dx du=dx v=(1/3)tan(3x) [xtan(3x)]/3  integral of(1/3)tan(3x)dx  (1/3)[ln(sec(3x))/3]  [ln(sec(3x))/9]

Hello! I really don't think I am understanding my calc hw. Please help me fix my errors. Thank you! 1. integral from 0 to pi/4 of (tanx^2)(secx^4)dx It says u = tan x to substitute So if I use u = tan x, then my du = secx^2 then I have integral of (u^2)

h t t p : / / i m g 4 0 . i m a g e s h a c k . u s / c o n t e n t . p h p ? p a g e = d o n e& l = i m g 4 0 / 1 4 8 / a s d a s d y e . j p g & v i a = m u p l o a d (def of tan theta = a^1 o)a = o opposite = adjacent tan theta written with respect to

Show that if A, B, and C are the angles of an acute triangle, then tan A + tan B + tan C = tan A tan B tan C. I tried drawing perpendiculars and stuff but it doesn't seem to work? For me, the trig identities don't seem to plug in as well. Help is

integral of Sec[2x]Tan[2x] i know u is sec 2x du=2sec2xtan2x dx what would i have to multiply with du so it would equal tan 2x dx? if my question is confusing, then here's another example of what i'm talking about: integral of (3x2)^30 dx u=3x2 du=3 dx

Calculate the following integral: ∫ sec^4 (3x)/ tan^3 (3x) dx For this one, can I bring up the tan to tan^3?

Hello im trying to integrate tan^3 dx i have solved out the whole thing but it doesnt match up with the solution.. this is what i did: first i broke it up into: integral tan^2x (tanx) dx integral (sec^2x1)(tanx) dx then i did a u substitution u = secx du

Hello im trying to integrate tan^3 dx i have solved out the whole thing but it doesnt match up with the solution.. this is what i did: first i broke it up into: integral tan^2x (tanx) dx integral (sec^2x1)(tanx) dx then i did a u substitution u = secx du

find d/dx (integral from 2 to x^4) tan(x^2) dx tan(x^4)^2*4x^3 MY ANSWER

Not sure if it is right, I have check with the answer in the book and a few integral calculators but they seem to get a different answer ∫ sec^3(x)tan^3(x) dx ∫ sec^3(x)tan(x)(sec^2(x)1) dx ∫ tan(x)sec(x)[sec^4(x)sec^2(x)] dx ∫


tan 15° tan (45°30°) (tan 45°  tan 30°)/1+ tan 45°tan30° (1√3/3)/(1+1√3/3) then i donnt what to do/ chancel out. Can someone finish it if i didn't get it wrong. thanks in advance

"Evaluate the following indefinite integral using integration by parts: *integral sign* tan^1(x) dx" I let u = tan^1(x) and dv = dx. Is that right?

Hello, everyone: I am working on finding the exact values of angles that are less common and are therefor not found easily on the Unit Circle (at least, they are not labeled). For example, the problem I am asking about is: 10) Find the exact values of the

Calc length of arc of y=ln(x) from x=1 to x=2  So far: Definite Integral over x=(1,2) of sqrt(1 + 1/x) dx 1/x = tan^2 t x = 1/tan^2 t sqrt(1+1/x) = sqrt(1+tan^2 t) = sec t dx = 2 * tan^3 t * sec^2 t dt Integrate over x=(1,2): sec^3 t / tan^3 t dt

Find the exact value of the given expression. (tan17pi/12  tan pi/4) / (1+tan 17pi/12 tan pi/4) can you help with this? i got the answer 1/ squareroot 3

can anyone tell me if tan1x= 1 over tan x? No. They are different. tan^1 (x) is a frequently used way of writing arctan x, which means the angle whose tangent is x. tan (tan^1 x) = x

sec^2xcotxcotx=tanx (1/cos)^2 times (1/tan)(1/tan)=tan (1/cos^2) times (2/tan)=tan (2/cos^2tan)times tan=tan(tan) sq. root of (2/cos^2)= sq. root of (tan^2) sq. root of (2i)/cos=tan I'm not sure if I did this right. If I didn't, can you show me the

Can someone check my answers so far? 1. Find the exact value of tan(3ð) Answer: tan(3ð) = tan (3 + 2ð) = tan (ð) = tan (ð + 2ð) = tan(ð) = sin (ð)/cos(ð) = (0)/(1) = 0 Exact value is 0 2. Solve triangleABC is C = 90 degrees, B = 20 degrees,

Hi I need some assistance on this problem find the exact value do not use a calculator cot[(5pi)/12] Here is my attempt RT = square root pi = 3.14...

Calculate definite integral of dx/(x^4 * sqrt(x^2 + 3)) Over (1,3) I start with the substitution x = sqrt(3)*tan t so: sqrt(x^2 + 3) = sqrt(3) * sec t dx = sqrt(3) * sec^2 t dt x^4 = 9 * tan^4 t The integral simplifies to: = dt/(tan^3 t * sin t) How do I


Use an addition or subtraction formula to write the expression as a trigonometric function of one number, and then find its exact value. (tan(π/2)+tan((2π)/3)) / (1tan(π/2)tan((2π)/3)) Answer=__________ Exact value answer=_________

∫ tan^2 (x) sec^4 (x) dx ∫ [tan^2 (t) + tan^4 (t)] dt ∫ [1tan^2 (x)] / [sec^2 (x)] dx Trigonometric integral Please show steps so I can understand!

Use an Addition or Subtraction Formula to write the expression as a trigonometric function of one number. (tan(74°)tan(14°))/(1+tan(74°)tan(14°)) =__________ Find it's exact value (not decimal form) =__________ Note: A similar question(but with

Find the first derivative for the following functions 1) f(x) = sin(cos^2x) cos(sin^3x) 2) f(x) = ( tan 2x  tan x ) / ( 1 + tan x tan 2x ) 3) f(x) = sin { tan ( sqrt x^3 + 6 ) } 4) f(x) = {sec^2(100x)  tan^2(100x)} / x

If limit as delta x approaches 0 of tan(0+Δx)tan(0)/Δx =1 which of the following is false: d/dx [tanx]=1 the slope of y = tan(x) at x = 0 is 1 y = tan(x) is continuous at x = 0 y = tan(x) is differentiable at x = 0

Integrate: dx/sqrt(x^29) Answer: ln(x + sqrt(x^2  9)) + C I'm getting the wrong answer. Where am I going wrong: Substitute: x = 3 * sec t sqrt(x^2  9) = sqrt(3) * tan t dx = sqrt(3) * sec t * tan t Integral simplifies to: sec t dt Integrates to: lnsec

I need help on finding the local linear approximation of tan 62 degree. i got 1.77859292096 can someone check if i got it right? tan(60+2) = (tan 60+ tan 2) / (1 tan60 tan 2) But tan2 appx = sin 2deg = sin2PI/180= 2PI/180 tan 62=(tan 60+2PI/180) / (1

Evaluate the indefinite integral integral sec(t/2) dt= a)ln sec t +tan t +C b)ln sec (t/2) +tan (t/2) +C c)2tan^2 (t/2)+C d)2ln cos(t/2) +C e)2ln sec (t/2)+tan (t/2) +C

Write the expression as the sine, cosine, tangent of an angle (in radians.) (tan(pi/5)tan(pi/3))/(1+tan(pi/5)tan(pi/3))

Determine all triangles ABC for which tan(AB)+tan(BC)+tan(CA)=0. There's a hint: "Can you relate AB to BC and CA?" Should I apply the tangent difference formula (tan(xy))? Help would be appreciated, thanks.


Prove that tan(A+B+C) =(tan A + tan B tan C  tanAtanBtanC) ÷(1tanAtanBtanBtanCtanAtanC)

Evaluate the limit as h > 0 of: [tan (pi/6 + h)  tan(pi/6)]/h I thought the answer was √3/3, or tan(pi/6, but apparently that is wrong, any tips here?

So I am suppose to evaulate this problem y=tan^4(2x) and I am confused. my friend did this : 3 tan ^4 (2x) d sec^ 2x (2x)= 6 tan ^4 (2x) d sec^2 (2x) She says it's right but what confuses me is she deriving the 4 and made it a three? I did the problem like

Write the given expression as the tangent of an angle. tan 7x − tan x/ 1 + tan 7x tan x

Which trigonometric functions are equivalent to tan(theta)? Select all that apply. (2 answers) a. tan(theta)**** b. tan(theta) c. tan(theta) d. tan(theta+2pi)**** e. tan(theta+pi/2)

y= cube root (1+tan(t)) OR y= (1+tan(t))^(1/3) The answer I got was y'= (1/3)(1+tan(t))^(2/3)*(sec^2(t)) Is this correct?

How do I derive the integration reduction formula for tangent? Integral of (tan x)^n dx = ... I can do the derivations for sin/cosine, but I'm getting stuck on tan. Thanks!

What is the integral of 1tan^2theta from 0 to pi/3? Using the identity tan^2theta=sec^2theta1, I got my answer to be 2pi/3  sqrt*3 Can someone verify this for me please?

For each of the following determine whether or not it is an identity and prove your result. a. cos(x)sec(x)sin^2(x)=cos^2(x) b. tan(x+(pi/4))= (tan(x)+1)/(1tan(x)) c. (cos(x+y))/(cos(xy))= (1tan(x)tan(y))/(1+tan(x)tan(y)) d.

Find the exact value of thet following trigonometric functions: tan(5pi/6) tan(7pi/6) tan(11pi/6)


Find the smallest positive value of x(in degree) for which tan(x+100°)=tan(x+50°)*tanx*tan(x50°)

Given tan 0 = 5/12 determine the possible values of 0 to the nearest hundreth So what I did is find that tan1(5/12) equals 0.35. Then because tan can be negative in quadrants 1 and 3 subtract 0.35 from pi and then from 2pi. So I got 2.79 and 5.933 I'm

integral of tan^5(2x)sec^3/2(2x)dx Y^11/11  2/7Y^7 + 1/3Y^3 where: Y = (1 + u^2)^(1/4) and: u = tan(2x)

could someone please tell me what the other angle identity of tan is? Thank you much! "...other angle identity of tan x"? in relation to which "other" ? there are many identities dealing with the tangent. The most important is probably tan x = sin x/cos x

I'm having a little trouble with this problem...it would be great if you could point out where I'm going wrong. Hours of Daylight as a Function of Latitude. Let S(x) be the number of sunlight hours on a cloudless June 21st, as a function of latitude, x,

find dy/dx y=ln (secx + tanx) Let u= secx + tan x dy/dx= 1/u * du/dx now, put the derivative of d secx/dx + dtanx/dx in. You may have some challenging algebra to simplify it. Use the chain rule. Let y(u) = ln u u(x) = sec x + tan x dy/dx = dy/du*du/dx

Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v  u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos v + cos v sin u. cos (v 

Ok, I have two questions first: 1. I'm asked to find a cartesian equation for the polar graph of this polar equation: r^2 = sin(2(theta)) The answer I got was (x^2+y^2)^2/(2xy) = 2 Is this the correct way to express it? 2. I need to find the cartesian

if you can't help me with my first question hopw you can help me with this one. sec(x)/csc(x)=tan(x) thanx to anyone who can help From the definition of the sec and csc functions, and the tan function, sec(x)/csc(x) = sin(x)/cos(x) = tan(x) However,

2. solve cos 2x3sin x cos 2x=0 for the principal values to two decimal places. 3. solve tan^2 + tan x1= 0 for the principal values to two decimal places. 4. Prove that tan^2(x) 1 + cos^2(x) = tan^2(x) sin^2 (x). 5.Prove that tan(x) sin(x) + cos(x)=


Use a sum or difference identity to find the exact value. tan 25deg + tan 5deg / 1tan 25deg tan 5deg

Find the integral of x/cos^2(x^2)dx The answer is 1/2 tan (x^2) + C. Where does the 1/2 come from?

1.Solve tan^2x + tan x – 1 = 0 for the principal value(s) to two decimal places. 6.Prove that tan y cos^2 y + sin^2y/sin y = cos y + sin y 10.Prove that 1+tanθ/1tanθ = sec^2θ+2tanθ/1tan^2θ 17.Prove that sin^2wcos^2w/tan w sin

Since sec^2θ  1 = tan^2θ I know this is trivial, but I want to make sure I'm doing this right before I apply it to the integral I'm trying to solve... If I have some constant a, (a^2secθ)^2  (a^2)^2 If I wanted to change this to tan would it be:

a billboard 20ft tall is located on top of a building with its lower edge 60 ft above the level of a viewers eye. how far from a point directly below the sign should a viewer stand to maximize the angle (theta) between the lines of sight of the top and

1. Find the indefinite integral. Indefinite integral tan^3(pix/7)sec^2(pix/7)dx 2. Find the indefinite integral by making the substitution x=3tan(theta). Indefinite integral x*sqrt(9+x^2)dx 3. Find the indefinite integral. Indefinite integral

Sum and diffence formula Finding Exact value of Tan 105Tan 10)15)/1+ tan(105)Tan(15)

Let F(x, y, z) = z tan−1(y^2)i + z^3 ln(x^2 + 10)j + zk. Find the flux of F across S, the part of the paraboloid x^2 + y^2 + z = 19 that lies above the plane z = 3 and is oriented upward. Thanks! STudying for a quiz... *tan−1 is tan^(1)

Let F(x, y, z) = z tan−1(y^2)i + z^3 ln(x^2 + 10)j + zk. Find the flux of F across S, the part of the paraboloid x^2 + y^2 + z = 19 that lies above the plane z = 3 and is oriented upward. Thanks! STudying for a quiz... *tan−1 is tan^(1)

Can someone tell me what the radian value is for this number and how they got it? Thanks On a graph plot the point A(2,1) Let C(2,0) be the point on the xaxis x is the angle AOB and tan x = 1/2 Since angle x is not one of the special angles, you will have


Find the value of tan[tan^1 1/2] My answer is 1/2. Is this correct?

1)tan 5 degrees + tan 25 degrees / 1 tan 5 degrees tan 25 degrees = sqrt 3 / 3 Am I correct? 2) Complete the identity tan^2Q  3sinQtanQsecQ = 2tan^2Q Correct?

I have two problems I am stuck on, if you could show me how to solve the problems it would be much appreciated. 1) Find sin 2x, cos 2x, and tan 2x from the given information. tan x = − 1/6, cos x > 0 sin 2x = cos 2x = tan 2x = 2) Find sin 2x, cos

please help me. use trig. identities to find the exact value. tan 25° + tan 5° / 1 tan 25° tan 5°

please help me. use trig. identities to find the exact value. tan 25° + tan 5° / 1 tan 25° tan 5°

Show that tan^3(x)8(tan^2(x))+17(tan(x)08 = 0 has a root in [0.5, 0.6]. Apply the Bisection Method twice to find an interval of length 0.025 containing this root. I have NO idea where to go with that equals 0 part...I could figure it without it but that

How can the right equation be converted into the answer on the left? (sec^4x)(tan^2x)=(tan^2x + tan^4x)sec^2x

I have a test soon.. and I really need to know how to do this problem.. please help!!! lim as x>0 sin^2(x)/tan(x^2) the answer is 1, but I have no clue how to get that! Use series expansions. Look up Taylor expansion on google and study that topic

There's a question on my homework that says: Tan(x)=10/17. I an not sure how to solve the Tan(x) part. Am I supposed to do Tan1(x)? Could someone help please?? Thanks!

Below are the 5 problems which I had trouble in. I can't seem to get the answer in the back of the book. Thanks for the help! lim (thetapi/2)sec(theta) theta>pi/2 Answer: 1 I am not sure what to do here. lim (tan(theta))^(theta) theta>0+ Answer:1


Please check for me .Thanks A building cast a shadow 23 feet long when the angle of the sun is 52 degree. Find the height of the building to the nearest integer. Let X be the height of the building Tan 52 degree =x/23 23 tan 56 =294feet x = 23 tan 52 =

what is the integral of {sin(x)tan(x)} i tried turning tan(x) into sin(x)/cos(x) then doing u substitutions but i always have an extra sin(x) left. can you help me pleeeease

A=170 degree then prove that Tan A/2=1rot(1+Tan^2 A)/Tan A

1)tan^2x=3 ? 2)tan x sin^2x=tan x ? 3)2cosx sinx  cos x=0

1)tan 5 degrees + tan 25 degrees / 1 tan 5 degrees tan 25 degrees = sqrt 3 / 3 Am I correct? 2) Complete the identity tan^2Q  3sinQtanQsecQ = 2tan^2Q Correct?

1)tan 5 degrees + tan 25 degrees / 1 tan 5 degrees tan 25 degrees = sqrt 3 / 3 Am I correct? 2) Complete the identity tan^2Q  3sinQtanQsecQ = 2tan^2Q Correct?

Determine each tangent ratio to four decimal places using a calculator. A) tan 74 degrees B) tan 45 degrees C) tan 60 degrees D) tan 89 degrees How do I figure this out !?

The directions are write the expression as the sine, coine, or tangent of an angle. (tan 25 degrees + tan 10 degrees) / [1  [(tan 25 degrees)(tan 10 degrees)]]

Find Indefinite Integral of dx/(x(x^4+1)). I think that Im complicating it too much. I moved the dx out making it 1/(x(x^4+1))dx than 1/(x^5+x)dx. I think i have to use formula indef integral of dx/(x^2+a^2) = 1/a(tan^1(x/a)) but im stuck i don't know

Evaluate tan(cos^(1)(ã3/2 )+tan^(1)(ã3/3))? arccos(sqrt(3) / 2) = pi / 6 arctan(sqrt(3) / 3) = pi / 6 tan(pi / 6 + pi / 6) = tan(pi / 3) = sqrt(3).


2 given the curve is described by the equation r=3cos ¥è, find the angle that the tangent line makes with the radius vector when ¥è=120¨¬. A. 30¨¬ B. 45¨¬ C. 60¨¬ D. 90¨¬ not sure A or D 2.) which of the following represents dy/dx when

Given that tan 45=1, use tan(x+y) to show that tan 22.5= /2  1. (/is square root sign

Solve the following trig equations. give all the positive values of the angle between 0 degrees and 360 degrees that will satisfy each. give any approximate value to the nearest minute only. a.) ( I got confused doing this 1 can you help me with it.) 3 sin

The radar stations A and B, separated by the distance a = 500 m, track the plane C by recording the angles α and β at 1second intervals, If three successive readings are shown in the table,calculate the speed v of the plane and the climb angle y

Could someone explain the steps for this question. Simplify the expression tan(5pi/8)tan(3pi/8)/1+tan(5pi/8)tan(3pi/8)

Am I allowed to do this? for the integral of ∫ sec^4 (3x)/ tan^3 (3x) dx I change it to ∫ sec^4 (3x) tan^3 (3x) From here I use the rule for trigonometry functions.

Use the exact values of the sin, cos and tan of pi/3 and pi/6, and the symmetry of the graphs of sin, cos and tan, to find the exact values of si pi/6, cos 5/3pi and tan 4pi/3. I have found the answers to the first three using the special tables

sqrt(1+tan x)sqrt(1+sin x) lim all divided by x^3 x>0 Use that Sqrt[1+x] = 1+ 1/2 x + 1/2 (1/2)/2 x^2 + 1/2(1/2)(3/2)/6 x^3 + O(x^4) You can thus write the numerator as: 1/2 [tan(x)  sin(x)]  1/8 [tan^2(x)  sin^2(x)] + 1/16 [tan^3(x) 

Derivative of √ tan x is, Answer: A. sin 2 x/ √ tan x B. cos 2 x/ 2 √ tan x C. sec 2 x D. None of these

Please review and tell me if i did something wrong. Find the following functions correct to five decimal places: a. sin 22degrees 43' b. cos 44degrees 56' c. sin 49degrees 17' d. tan 11degrees 37' e. sin 79degrees 23'30' f. cot 19degrees 0' 25'' g. tan


Use the fundamental identities to simplify the expression: cot beta sec beta I used 1+tan^2u=secu since cot is the inverse of tan. I flipped the tangent, then so it was 1+ (1/tan). But the book's answer is the cosecant of beta. Where did this come from??

Let F(x)= the integral from 0 to 2x of tan(t^2) dt. Use your calculator to find F″(1) By applying the fundamental theorem of calculus, I got the derivative of the integral (F'(x)) to be 2tan(2x^2) When I take the derivative to find F''(x) I get 8x

Proving identities: 1) 1+ 1/tan^2x = 1/sin^2x 2) 2sin^2 x1 = sin^2x  cos^2x 3) 1/cosx  cosx = sin x tan x 4) sin x + tan x =tan x (1+cos x) 5) 1/1sin^2x= 1+tan^2 x How in the world do I prove this...please help... I appreciateyour time thankyou soo

I keep trying to solve this by factring but I can't seem to get! It's so frustrating! The problem is 1+tan2x=0 This is how I worked the problem before: 1+((2tanx)/(1tan^2x))=0 ((2tanx)/(1tan^2x))= 1 Then I cross multiplied to get: 1 + tan^2x = 2 tanx

If tan^1(x^2y^2/x^2+y^2) = a, prove that dy/dx = x(1  tan a)/y(1 + tan a) Please solve!!!